# An object with a mass of 2 kg is traveling at 14 m/s. If the object is accelerated by a force of f(x) = x+xe^x  over x in [0, 9], where x is in meters, what is the impulse at x = 8?

Mar 24, 2016

1

#### Explanation:

The definition of impulse is

$I = F \cdot \Delta t =$

and using the 2nd law of Newton $F = m \cdot a$ it is equivalent to the variation of momentum,

I=Deltap = m*Deltav = m*(v_f -v_0) ; m=2kg ; v_0=14m/s

so, we need to calculate the speed at the position x=8; v_f

By the 2nd law of Newton $F = m \cdot a \implies F \left(x\right) = m \cdot \frac{\mathrm{dv}}{\mathrm{dt}}$

$F \left(x\right) = m \cdot \frac{\mathrm{dv} \left(x\right)}{\mathrm{dx}} \cdot \frac{\mathrm{dx}}{\mathrm{dt}} = m \cdot \frac{\mathrm{dv} \left(x\right)}{\mathrm{dx}} \cdot v \left(x\right)$

${\int}_{0}^{8} F \left(x\right) \mathrm{dx} = m \cdot \left({v}_{f}^{2} - {v}_{0}^{2}\right)$

${v}_{f} = \sqrt{{v}_{0}^{2} + \frac{1}{m} \cdot {\int}_{0}^{8} F \left(x\right) \mathrm{dx}}$

thus, the impulse is

$I = m \cdot \left(\sqrt{{v}_{0}^{2} + \frac{1}{m} \cdot {\int}_{0}^{8} F \left(x\right) \mathrm{dx}} - {v}_{0}\right) = 262.5 \cdot k g \cdot \frac{m}{s}$